I took the following definition from Milne's Fields and Galois Theory (page 69):
The part I underlined is the one giving me trouble. In particular, I would like to know why a $G$-module by that definition is the same as giving a homomorphism $f: G \to \operatorname{Aut}(M)$.
- Criterion (a) ensures that $f$ maps to an endomorphism over $M$.
- Criterion (b) ensures that $f$ is indeed a homomorphism.
- Now I think that Criterion (c) guarantees that $f$ maps to an automorphism but I am not sure if this explanation is sufficient: We have $f_{1_G} = f_{\sigma \sigma^{-1}} = f_\sigma \circ f_{\sigma^{-1}}$ for any $\sigma \in G$ (second equality follows from (b)). Criterion (c) should guarantee that $f_\sigma^{-1} = f_{\sigma^{-1}}$. Since $f_\sigma$ is a endomorphism (cf. my first point) which has an inverse map, it must be an automorphism.
Is this argumentation correct?
The elements of $G$ are each assigned to a function acting on $M$; the map $f\colon G\to \operatorname{Aut}(M)$ tells you the assignment.
In this language, (a) says that $f(g)(m+m')=f(g)(m)+f(g)(m')$, i.e., that $f(g)$ is a module homomorphism (endomorphism) of $M$.
Now (b) says that $f(gh)(m)=f(g)(f(h)(m))$, so as functions $M\to M$, we have $f(gh)=f(g)f(h)$. Really, this says that $f$ is a monoid homomorphism from $G$ to $\operatorname{End}(M)$.
Together, note that $f(g^{-1})f(g)=f(g)(f(g^{-1}))=f(1)$, so (c) guarantees that $f(g)$ is always invertible. Thus, the assignment $g\mapsto f(g)$ is really a map $G\to\operatorname{Aut}(M)$, and in fact $f$ is a group homomorphism.
Now its easy to see that any group homomorphism $f\colon G\to \operatorname{Aut}(M)$ satisfies (a), (b), and (c), and these processes are inverse, so a group action of $G$ on $M$ is naturally equivalent to a homomorphism $G\to \operatorname{M}$.